Show that $s\longmapsto \int_a^b g(s,u)du$ is holomorphic.

Question

Let $U\subset \mathbb C$ an open and $a<b$ reals numbers. Let $g:U\times [a,b]\longrightarrow \mathbb C$ a continuous function s.t. $s\longmapsto g(s,u)$ is holomorphic for all $u\in [a,b]$. Show that $$s\longmapsto \int_a^b g(s,u)du,$$ is holomorphic.

Attempts

Let $h$ s.t. $s+h\in U$. We suppose WLOG that $0<|h|<1$. Since $s\longmapsto g(s,u)$ is holomorphic, there is $|c_h|<|h|$ s.t. $$\left|\frac{|g(s+h,u)-g(s,u)|}{h} \right|\leq \frac{\partial g}{\partial x}(c_h,u)\leq \frac{\partial g}{\partial x}(c,u)\in L^1(a,b),$$ where $c=\max_{k\in [s,s+1]}g(k,u)$.

Using dominated convergence theorem, the claim follow.

Does it work ?

Answer 1

• If $f_n $ is holomorphic on $U$ and $f_n \to f$ converges uniformly on $U$ then $f$ is holomorphic on $U$. Proof : for any closed contour $\gamma \subset U$ : $\int_\gamma f(z)dz = \int_\gamma \lim_{n \to \infty} f_n(z)dz= \lim_{n \to \infty} \int_\gamma f_n(z)dz= 0$. Thus $F(z) = \int_a^z f(z)dz$ is well-defined and holomorphic, and by the holomorphic $\implies$ analytic theorem, $f(z) = F'(z)$ is analytic and holomorphic too.

• Let $h_n(z) = \frac{1}{n}\sum_{m=1}^n g(z,a+\frac{b-a}{n})$ the sequence of Riemann sums of $h(z) = \int_a^b g(z,u)du$. Clearly $h_n$ is holomorphic. Since $g(z,u)$ is (locally) uniformly continuous, $h_n$ converges (locally) uniformly to $h$, thus $h$ is holomorphic.